Answer:
<u>Inelastic collision:</u>
A collision in which there is a loss of Kinetic Energy due to internal friction of the bodies colliding.
<u>Characteristics of an inelastic collision:</u>
- <em>the momentum of the system is conserved</em>
- <em>the momentum of the system is conservedloss of kinetic energy</em><u> </u>
<em>I</em><em>n</em><em> </em><em>a perfectly elastic collision</em><em>, the two bodies </em><em>that</em><em> </em><em>collide with each other stick together.</em>
<u>Elastic </u><u>collision</u><u>:</u>
A collision in which the kinetic energy of the two bodies, before and after the collision, remains the same.
<u>Characteristic</u><u>s</u><u> </u><u>of</u><u> </u><u>elastic</u><u> </u><u>collision</u><u>:</u>
- <em>the</em><em> </em><em>momentum</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>system</em><em> </em><em>is</em><em> </em><em>conserved</em>
- <em>no</em><em> </em><em>loss</em><em> </em><em>o</em><em>f</em><em> </em><em>kinetic</em><em> </em><em>energy</em>
In everyday life, no collision is perfectly elastic.
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ANSWER:
<u>Given examples:</u>
- Two cars colliding with each other form an example of inelastic collision.
<u>Reason:</u>
<em>(</em><em>T</em><em>hey</em><em> </em><em>lose</em><em> </em><em>kinetic</em><em> </em><em>energy</em><em> </em><em>and</em><em> </em><em>come</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>stop</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>collision</em><em>.</em><em>)</em>
- A ball bouncing after colliding with a surface is an example of elastic collision
<u>Reason:</u>
<em>(a very less amount of kinetic energy is lost)</em>
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
Answer:
W = - 5.01 10¹⁰ J
Explanation:
Work is defined by the expression
W = ∫ F.dr
Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product
W = ∫ F dr
W = G m₁m₂ ∫ 1 /r² dr
W = G m₁ m₂2(-1 / r)
We evaluate between the lower limits r = Re and upper r = ∞
W = G m₁m₂ (-1 / Re + 1 / ∞)
W = - G m₁ m₂ / Re
Let's calculate
W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶
W = - 5.01 10¹⁰ J
<span>Electrical discharge from a charged object
is your answer</span>
